Polynomial splittings of Ozsvath and Szabo's dinvariant
Abstract.
For any rational homology 3sphere and one of its spinstructures, Ozsváth and Szabó defined a topological invariant, called invariant. Given a knot in the 3sphere, the invariants associated with the primepowerfold branched covers of the knot, obstruct the smooth sliceness of the knot. These invariants bear some structural resemblances to CassonGordon invariants, which obstruct the topological sliceness of a knot. SeGoo Kim found a polynomial splitting property for CassonGordon invariants. In this paper, we show a similar result for Ozsváth and Szabó’s invariants. We give an application of the result.
Key words and phrases:
Alexander polynomial, knot concordance, Ozsváth and Szabó’s invariant.2010 Mathematics Subject Classification:
Primary 57M27 57M251. Introduction
We work in smooth category, and all manifolds are supposed to be smooth unless stated otherwise. An oriented knot in the 3sphere is said to be slice if there is a smoothly embedded 2disk in the 4ball satisfying . Here is called a slice disk of . A pair of knots and are said to be smoothly concordant (which is denoted by ) if is slice where is the mirror image of with reversed orientation. Smooth concordance is an equivalence relation among knots and the set of equivalence classes becomes an abelian group under the operation of connected sum. The group is called the knot concordance group and denoted by . Slice knots represent the zero element in .
A knot is said to be ribbon if bounds a smoothly immersed 2disc in which has the property that the preimage of each component of selfintersection consists of a properly embedded arc in the disc and an arc embedded in the interior of the disc. It is obvious that each ribbon knot is smoothly slice.
To study the group structure of , there are two basic questions to consider. First, given a knot, we need to figure out what the order of the knot is in . Second, given several knots , we want to know if they are linearly independent or not in . Finite order elements in are called torsion elements. As for the first question, the only known torsion in is 2torsion, which comes from negative amphicheiral knots. Some invariants such as signature, Rasmussen invariant and Ozsváth and Szabó’s invariant, induce homomorphisms from to the group of integers. If a knot is a torsion element, it has vanishing value on such invariants.
For the independence problem in , there is a systematic way to study it by considering the relative primeness of the Alexander polynomials. Levine [9] first showed that if the connected sum of two knots with relatively prime Alexander polynomials has vanishing Levine obstructions, then so do both knots. Kim in [7] showed that the CassonGordonGilmer obstruction splits in the same manner. In [8], a similar splitting property was proved for von Neumann invariants associated with certain metabelian representations.
In this paper, we study a similar splitting property for Ozsváth and Szabó’s invariants, and apply this property to study the independence problem in .
Given a 3manifold and one of its torsion spinstructures , the invariant is defined for by Ozsváth and Szabó in [11]. Let be a knot in , and be the fold cyclic branched cover of along with for some prime number . Then is a rational homology 3sphere. Therefore we can consider the invariant of for any of its spinstructures.
If is a slice knot, let be the fold branched cover of along a slice disk of . It is known that is a rational homology 4ball whose boundary is . As studied in [11] and reformulated in many papers such as [3, 5], many of the invariants for must vanish (see Theorem 2.2).
For any spinstructure over , define
where is a special spinstructure being discussed in Section 2. The first homology group acts freely and transitively on the set of spinstructures . Given an element and an element , let denote the resulting element under the group action. We prove the following theorem.
Theorem 1.1.
Let and be two knots whose Alexander polynomials are relatively prime in . Suppose further that at least one of and has nonsingular Seifert form. Then the following hold.

If is slice, then for all but finitely many prime numbers there exists a subgroup satisfying such that for any and , where can be any power of .

If is ribbon, the conclusion holds for any prime number .
For the invariant defined in [3], analogous properties can be proved by using the same argument. Furthermore, for invariants and which are defined similarly as and in [3], we have Theorem 2.7.
As an application of the results above, we show the following property, which has been know before [7].
Proposition 1.2.
Let be the twist knot. Excluding the unknot, (which is the figure8 knot) and (which is Stevedore’s knot), no nontrivial linear combinations of twist knots are ribbon knots.
Acknowledgements
I would like to thank Charles Livingston and SeGoo Kim for helpful discussion. My special thanks are due to Livingston for proposing to me the question concerning Lemma 2.3, and for telling me a concise proof that one can choose a slice disk for a ribbon knot which bounds a handlebody in the 4ball.
The author was supported by GrantinAid for JSPS Fellows, and by Platform for Dynamic Approaches to Living System from the Ministry of Education, Culture, Sports, Science and Technology, Japan.
2. Proof of Theorem 1.1
2.1. Alexander polynomial, Seifert form and invariant
In this subsection, we review some backgrounds needed in the proof of Theorem 1.1. Let be a knot in with a Seifert surface of genus . Define the Seifert form as for any simple closed curves representing elements in , where denotes the map that pushes a class off in the positive normal direction. Fix a basis for and let be the Seifert matrix associated with this basis.
A Seifert form on is said to be nullconcordant if there exists a direct summand of so that and . Such a direct summand is called a metabolizer of .
A knot which has a nullconcordant Seifert form is called algebraically slice. Two knots and are said to be algebraically concordant if is algebraically slice. The set of the equivalence classes of knots under this relation becomes a group as well, called algebraic concordance group, and we denote it .
The following lemma is Lemma 3.1 in [7], which was refined from [6, Proposition 3]. Note that the definition of nullconcordance of a Seifert form in [7] is different from the one we are using, but the two definitions are equivalent.
Lemma 2.1 (Kervaire, Levine, Kim).
Given two knots and , let be a Seifert surface of , and be the Seifert form on for . Suppose the Alexander polynomials of and are relatively prime in , and either or is nonsingular. Then if is nullconcordant with a metabolizer , then is nullconcordant with metabolizer for .
From this lemma we see that, with the assumption in this lemma, if is algebraically slice, so are both and .
Let be defined as before for a given knot and a prime power . The homology group acts freely and transitively on the set . So a choice of one spinstructure gives a bijection between and . As discussed in Section 2 of [3], there exists a canonical spinstructure , which is uniquely characterised by and . For more details about the definition, please refer to [3] and [4]. If has no 2torsion, then is the unique spinstructure over . Under this , we can identify and , by sending to . One nice property of is that it is compatible with the connected sum of knots. Namely,
for two given knots and , where denotes the spinstructure of .
If the knot is slice, as before let be the fold branched cover of along a slice disk of . Consider the homomorphism induced by the inclusion map. Then it is known that a spinstructure over can be extended to if and only if has the form for some . As an application of the results in [11], the following theorem is known (see also [3]).
Theorem 2.2 (Ozsváth and Szabó).
If is slice, then for any .
2.2. Proof of Theorem 1.1
The order of is determined by the Alexander polynomial of . Precisely we have Fox’s formula
We prove the following lemma.
Lemma 2.3.
Given a knot and a prime number , let be the set of prime numbers with the property that if then there is certain integer such that divides the order of . Then is a finite set.
Proof.
Given a prime power , consider the resultant of and over the field of complex numbers . Then we have
where is the leading coefficient of . If divides the order of , then divides . Therefore in the field , which means and having a common root in the algebraic closure of . Since is never a root of , we see that and have a common root in the algebraic closure of .
Since , we have . We assume that the set is an infinite set. Since has only finitely many roots in the algebraic closure of , there must be two elements and in such that and have a common root in the algebraic closure of for some positive integers and . This contradicts Lemma 2.4. Therefore opposed to our assumption, the set is a finite set.
∎
Lemma 2.4.
If and are relatively prime integers greater than or equal to two, and can never have a common root in the algebraic closure of .
Proof.
It is enough to check that , which is nonzero in the algebraic closure of . In fact we have
(†) 
where
The first equation in († ‣ 2.2) is a basic property of the resultant of two polynomials. We prove the second equation by induction on . If , we assume that and subtract the ’s row from the ’s row in , where . Then we have
If , assume and subtract the ’s row from the ’s row, where . The resulting matrix is
Therefore .
(1) If , it follows from our induction that .
(2) If , then , which implies . Let be the integer for which for . Then we have
The condition implies that and . It follows from our induction that . ∎
Proof of Theorem 1.1.
(i) Choose a Seifert surface for the knot for . Let and . Then is a Seifert surface of . Suppose is a slice knot. Then bounds a 3manifold in , denoted by , where is a slice disk of in the 4ball . Consider the map induced by inclusion where is the torsion part of , and let . Then is a metabolizer of the Seifert form associated with . See Theorems 3.1.1 and 3.1.2 in [10] for detailed discussion.
Let denote sliced along , and denote sliced along . Considering the construction of and , we have the following commutative diagram.
The map and the maps in the vertical direction are induced by the inclusion maps. We have the following isomorphisms.
The second isomorphism follows from Alexander duality. In the same vein, we can establish the isomorphism . So we can replace the previous commutative diagram with the following diagram, which we denote by .
the maps and are induced by the inclusion maps.
We fix a basis for and let be the Seifert matrix of associated with this basis. Now we see there is a basis of which is naturally induced by . Under this basis, the map is represented by the matrix
where (See discussion before [2, Lemma 1] or [10, Theorem 6.2.2]). Here we abuse to denote both the map and the matrix. It is known that is a presentation matrix of . Define and for and respectively. Then .
Remember is the kernel of the map . Let . By Lemma 2.5, the order of is the square root of the order of . By Lemma 2.1, the metaboizer has a splitting where is a metabolizer of the Seifert form associated with , for . Then has a splitting where for . By Lemma 2.5 the order of is the square root of the order of for .
Since is finitely generated, there are only finitely many prime numbers dividing the order of , say they are elements in . We let where is the set described in Lemma 2.3, and then is again a finite set.
Remember that for some prime number and positive integer . Suppose the prime number is not in the set . In this case, we claim that
Since is the kernel of the map , so belongs to the torsion part of . Given , let . Then the order of divides the order of . On the other hand, the order of divides the order of . By the commutativity of the diagram (*) above, . If , the order of divides both the orders of and . So there exists an element dividing the order of . This conflicts with the choice of . Therefore , and our claim is proved.
By the commutativity of the diagram (), we have . Note that since is slice, the order of is the square root of the order of by Lemma 3 in [1]. Therefore and has the same order as finite groups, so . The invariants defined on are zero by Theorem 2.2.
We now show that the invariants of defined on are zero for both . For any element , the element is included in , so by the additivity of invariant we have
Here we abuse to denote the unique spinstructures discussed in Section 2.1 over , or . The value does not depend on the choice of , so we have
for any . The same fact can be proved for .
(ii) If is a ribbon knot, we can choose to be a handlebody, in which case is torsion free. Then the set is an empty set. Therefore the conclusion in holds for any prime power .
∎
In the rest of this section, we give a proof of the following lemma, which we cannot find a good reference for it.
Lemma 2.5.
Suppose is a metabolizer of the Seifert form for a Seifert surface of the knot . The order of is the square root of the order of for any prime power .
Seifert [12] proved that is a presentation matrix for with the set of generators . Namely we have an exact sequence
The map induces an isomorphism and is isomorphic to .
We prove that the order of is a square root of that of . The proof is similar to that of [2, Lemma 2]. As stated there, we may extend a basis of to a basis of . Under this basis, the matrices , and have the forms
where . By the invertibility of , we have . Therefore the order of is , while the order of is .
Proof of Lemma 2.5.
We show that is isomorphic to , which is isomorphic to . Following Lemma 1 in [2], there are integral determinant matrices and which can be written as block matrices whose blocks are polynomials in , such that has the form
where the stars mean some unspecified polynomials in . It is very easy to check that by the forms of and .
Next we show that
by using the properties of and . Remember that and are automorphisms of , so . Now we only need to show that induces an isomorphism between and . We show that
(‡) 
Choose an order for the elements in the basis of so that the elements in take the first positions. Remember that the blocks of are polynomials in . The form of tells us that under the reordered basis, the matrix and its inverse are matrices with the form
where stars are matrices. Since and its inverse are automorphisms, it is now easy to check that relations (‡ ‣ 2.2) hold.
∎
2.3. invariant, and
Let be the preimage of in . Considering as a knot in , Grigsby, Ruberman and Strle in [3] defined the invariant for and . This invariant satisfies the following property.
Theorem 2.6 (Grigsby, Ruberman and Strle).
If is slice, then for any .
Note that the proof of Theorem 1.1 only depends on the algebraic information carried on and Theorem 2.2, while does not depend on the definition of invariant. By replacing Theorem 2.2 with Theorem 2.6, we can prove exactly the same fact for invariant as we did for invariant in Theorem 1.1.
Grigsby, Ruberman and Strle in [3] also defined invariants and associated with the double branched cyclic cover of the knot . We can extend their definition naturally to the case of any branched cyclic cover with a prime power.
Suppose is a function on a finite abelian group and is a subgroup. Following [3] we let . Given a prime number , let be the set of all order subgroups of . Then we can define
and
Here we regard and as functions from to .
Given a function , we define by sending to . Let and be the invariants defined by taking and . We prove the following theorem:
Theorem 2.7.
Let be a positive prime number or 1. Suppose the Alexander polynomials of and are relatively prime in . Suppose further that at least one of and has nonsingular Seifert form.

If is a slice knot for some nonzero and , then for all but finitely many primes , the following holds: for , where is a power of .

If is a ribbon knot for some nonzero and , the conclusions above hold for any prime power .
3. Application
It is known in [9, Corollary 23] that the twist knot is

of infinite order in the algebraic concordance group if ;

algebraically slice if and is a square;

of finite order in otherwise.
Proof of Proposition 1.2.
The Alexander polynomial of the is
It is easy to check that for any two nontrivial twist knots, their Alexander polynomials are relatively prime in .
Note that each nontrival twist knot has nonsingular Seifert form. Excluding the unknot, 1twist knot and 2twist knot, suppose that there are two sets of positive integers and for which is a ribbon knot. Then
We consider the case when , namely the double branched covers of the twist knots.
Recall that has infinite order in the algebraic concordance group if . So each for is a positive integer.
Let be the 3manifold . Assume that and let be a prime number dividing . Then
Ozsváth and Szabó [11] provided a formula of the invariants for lens spaces, by which we have
for .
By calculation we have . So